Question

Какой объём занимает 1 кг кислорода при температуре 273 К и давление 8*10^5 Па?Молярная масса М=32*10^-3 кг/моль.

Answer 1

PV = mRT / M

V = mRT / MP

V = 1 * 8,3144 * 273 / 0,032 * 800000 = 0,089 м3

Answer 2

Calculation of the Volume of 1 kg of Oxygen

To calculate the volume of 1 kg of oxygen at a given temperature and pressure, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure in Pascal (Pa) - V is the volume in cubic meters (m³) - n is the number of moles of gas - R is the ideal gas constant (8.314 J/(mol·K)) - T is the temperature in Kelvin (K)

Given: - Mass of oxygen (m) = 1 kg - Molar mass of oxygen (M) = 32 * 10^-3 kg/mol - Temperature (T) = 273 K - Pressure (P) = 8 * 10^5 Pa

To find the volume (V), we need to calculate the number of moles (n) of oxygen using the mass and molar mass:

n = m / M

Substituting the given values: n = 1 kg / (32 * 10^-3 kg/mol) = 31.25 mol

Now, we can rearrange the ideal gas law equation to solve for volume (V):

V = nRT / P

Substituting the values: V = (31.25 mol) * (8.314 J/(mol·K)) * (273 K) / (8 * 10^5 Pa)

Calculating the volume using the given values and performing the calculation, we find that the volume of 1 kg of oxygen at a temperature of 273 K and a pressure of 8 * 10^5 Pa is approximately 0.109 m³.

Please note that the above calculation assumes ideal gas behavior and does not take into account any deviations from ideal gas behavior that may occur at high pressures or low temperatures.